Problem: Find $\lim_{x\to 0}\dfrac{\sin(x)}{\sin(2x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{2}$ (Choice B) B $1$ (Choice C) C $2$ (Choice D) D The limit doesn't exist
Explanation: Substituting $x=0$ into $\dfrac{\sin(x)}{\sin(2x)}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since our expression includes trigonometric functions, let's try to re-write it using factorization and trigonometric identities. Since we have $\sin(2x)$ in our expression, let's rewrite it using its double angle identity $\sin(2x)=2\sin(x)\cos(x)$. $\begin{aligned} &\phantom{=}\dfrac{\sin(x)}{\sin(2x)} \\\\ &=\dfrac{\sin(x)}{2\sin(x)\cos(x)} \gray{\sin(2x)\text{ identity}} \\\\ &=\dfrac{\cancel{\sin(x)}}{2\cos(x)\cancel{(\sin(x))}} \gray{\text{Cancel common factors}} \\\\ &=\dfrac{1}{2\cos(x)}\text{, for }x\neq \{...,-2\pi, -\pi, 0, \pi, 2\pi,...\} \end{aligned}$ This means that the two expressions have the same value for all $x $ -values (in their domains) except for $k\pi$ for any integer $k$, and specifically $0$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{\sin(x)}{\sin(2x)}=\dfrac{1}{2\cos(x)}$ for all $x$ -values in the interval $(-1,1)$ except for $x=0$. Therefore, $\lim_{x\to 0}\dfrac{\sin(x)}{\sin(2x)}=\lim_{x\to 0}\dfrac{1}{2\cos(x)}=\dfrac{1}{2}$. (The last limit was found using direct substitution.) In conclusion, $\lim_{x\to 0}\dfrac{\sin(x)}{\sin(2x)}=\dfrac{1}{2}$.